a.
\(\Leftrightarrow\left\{{}\begin{matrix}y=1-x\\x^2-2x+2y+2=0\end{matrix}\right.\)
\(\Rightarrow x^2-2x+2\left(1-x\right)+2=0\)
\(\Leftrightarrow x^2-4x+4=0\)
\(\Leftrightarrow x=2\Rightarrow y=-1\)
b.
\(\Leftrightarrow\left\{{}\begin{matrix}9x^2-9xy=18\\4x^2+2xy-2y^2=18\end{matrix}\right.\)
\(\Rightarrow5x^2-11xy+2y^2=0\)
\(\Leftrightarrow\left(5x-y\right)\left(x-2y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=5x\\y=\frac{1}{2}x\end{matrix}\right.\) thế vào pt đầu...
a) Ta có: \(\left\{{}\begin{matrix}x+y=1\\x^2-2x+2y+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1-y\\\left(1-y\right)^2-2\cdot\left(1-y\right)+2y+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1-y\\y^2-2y+1-2+2y+2y+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1-y\\y^2+2y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-y\\\left(y+1\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1-y\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-\left(-1\right)=2\\y=-1\end{matrix}\right.\)
Vậy: (x,y)=(2;-1)
